find the centre of mass of three particles at the vertices of an equilateral triangle. the masses of the particles are 100g, 150g and 200g respectively. each side of the triangle is 0.5cm long. if two vertices of the triangle are (0,0) and (0.5,0). what will be the third vertex.

**Case 1**:-

AB = BC = CA = 0.5 m

L = AB cos30°

*m* _{A} = 100 g, = 0.1 kg

*m* _{B} = 150 g, = 0.15 kg

*m* _{C} = 200 g, = 0.20 kg

So. C.O.M. = (X* _{cm} *, Y

_{CM}) = (0.278 m, 0.095 m)

**Case 2**:- assume the third vertex to be P(x,y) while the given other two vertices are A(0,0) and B(0.5,0) and since it is an equilateral triangle:

AP = BP --------(2)

using the distance formula[]we can get the desired result by substituting values in equation (2) and equating it.

**
**